What is the integral from (-1) to 3 of-|x-4|+2 ?

The integral from (-1) to 3 of-|x-4|+2 is -4

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integral from (-1) to 3 of-|x-4|+2

\int_{(- 1)}^{3} - ∣ x - 4 ∣ + 2 d x

- 4

Solution steps

\int_{(- 1)}^{3} - ∣ x - 4 ∣ + 2 d x

Eliminate Absolutes

= \int_{(- 1)}^{3} - (- x + 4) + 2 d x

Simplify

= \int_{(- 1)}^{3} x - 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{(- 1)}^{3} x d x - \int_{(- 1)}^{3} 2 d x

\int_{(- 1)}^{3} x d x = 4

\int_{(- 1)}^{3} 2 d x = 8

= 4 - 8

Simplify

= - 4

x^{2} sin (\frac{1}{x})

\frac{x}{x ^{2} + 81}

\frac{\partial}{\partial y} (x e^{x^{3} y})

\frac{d}{d x} ((x + 2) e^{x})

\frac{\partial}{\partial x} (\frac{- 2 x - y}{( x - y ) ^{4}})

What is the integral from (-1) to 3 of-|x-4|+2 ?
The integral from (-1) to 3 of-|x-4|+2 is -4